Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)

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- Oct 13th 2009, 10:05 AMCoda202Isomorphic
Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)

- Oct 13th 2009, 10:47 AMeeyore
I'll give it a shot. Assume is such an isomorphism. Say where . must satisfy

so

Thus, each integer, and in turn each rational, is mapped according to the identity mapping. Then where would the irrationals go? The homomorphism seems to be onto, but not one-to-one, so no bijection can exist. - Oct 13th 2009, 10:51 AMTaluivren
Hi,

here's my try:

sentence

holds in but not in . So they are not elementarily equivalent, therefore they can't be isomorphic. - Oct 13th 2009, 12:08 PMclic-clac
Another one: assume is a ring isomorphism, then .

But no rational is a squared root of 2, so cannot exist.

(assume belong to and then .

But this is impossible since is even and is odd (where is the greatest such that divides ))